Question

$\int \frac{e^x-1}{e^x+1}dx.$

• A. $\log (e^x+1)-\log e^x$
• B. $2\log (e^x-1)-\log e^x$
• C. $2\log (e^x+1)-\log e^x$
• D. $2\log (e^x+1)+\log e^x$

Answer 1

Answer: (C)

$2\log (e^x+1)-\log e^x$

Let $I=\int \frac{e^x-1}{e^x+1}dx=\int \frac{e^x(e^x-1)}{e^x(e^x+1)}dx$

Put $e^x=t\Rightarrow e^{x}dx=dt$

$I=\int \frac{t-1}{t(t+1)}dt=\int (\frac{2}{t+1}-\frac{1}{t})dt$

$=2\log (t+1)-\log t=2\log (e^x+1)-\log e^x$